Question: What is the extraneous solution to these equations? $\dfrac{x^2 + 6x}{x - 1} = \dfrac{-x + 8}{x - 1}$
Multiply both sides by $x - 1$ $ \dfrac{x^2 + 6x}{x - 1} (x - 1) = \dfrac{-x + 8}{x - 1} (x - 1)$ $ x^2 + 6x = -x + 8$ Subtract $-x + 8$ from both sides: $ x^2 + 6x - (-x + 8) = -x + 8 - (-x + 8)$ $ x^2 + 6x + x - 8 = 0$ $ x^2 + 7x - 8 = 0$ Factor the expression: $ (x - 1)(x + 8) = 0$ Therefore $x = 1$ or $x = -8$ At $x = 1$ , the denominator of the original expression is 0. Since the expression is undefined at $x = 1$, it is an extraneous solution.